A long straight wire AB carries current 10 A. A proton (P) travels with a speed $2×10^6 m/s$, parallel to the wire, at a distance 2 m from it in a direction opposite to the current as shown in figure. The force experienced by the proton is: $(μ_0=4π×10^{-7}NA^{-2})$

$3.2×10^{-19}N$

The correct answer is Option (3) → $3.2×10^{-19}N$

The magnetic field (B) at a distance (r) from a straight wire carrying current (I) is:

$B=\frac{μ_0I}{2πr}$ [formula]

$μ_0=4π×10^{-7}$

$I=10A$

$r=2m$

$B=\frac{4π×10^{-7}×10}{2π×2}=1×10^{-6}T$

Now,

Magnetic force, $F=qvB\sin θ$

where,

q, charge of proton = $1.6×10^{-19}C$

v, velocity of proton = $2×10^6m/s$

B, Magnetic field = $1×10^{-6}T$

θ, ($\vec v$ is perpendicular to $\vec B$) = 90°

$F=1.6×10^{-19}×2×10^6×1×10^{-6}×\sin 90°$

$=3.2×10^{-19}N$